-2

Let $a$, $b$ be a positive integers such that $(a+b)\mid ab$. Prove that $\gcd(a,b)\geq\sqrt{{a}+{b}}$. I have no idea how to approach this problem. Any hints or solutions are appreciated.

Bernard
  • 175,478
somerndguy
  • 521
  • 1
    Hint: By the Theorem in the dupe $\ a+b\mid ab\iff a+b\mid (a,b)^2\Rightarrow a+b \le (a,b)^2\ $ – Bill Dubuque Dec 07 '20 at 17:29
  • By $(a,b)$ you mean $gcd(a,b)$? If so then let $gcd(a,b)=d$ and by your last inequality we have $a+b/leq d^2$ and as both sides are positive we have $\sqrt{{a}+{b}}\leq d$ which is what was to be proved. Very nice and simple observation thank you. – somerndguy Dec 08 '20 at 11:04
  • Yes. that's correct. – Bill Dubuque Dec 08 '20 at 11:43

2 Answers2

2

Let $a=d\cdot x$ and $b=d\cdot y$ with $x$ and $y$ coprime (i.e. $\gcd(x,y)=1$). This means that $\gcd(a,b)=d$. So your question translates to $d\geq \sqrt{d(x+y})$ knowing that $d(x+y)|d^2xy$.

In other words, if $x+y|dxy$ then $d\geq x+y$.

Suppose $k|xy$ and $k|x+y$. Then $k|(x+y)\cdot x=xy+x^2$, so $k$ also divides $x^2$. Analogously, $k(x+y)\cdot y$ so $k|y^2$. But $x$ and $y$ are coprime, so $x^2$ and $y^2$ are coprime, so $K=1$, meaning that $x+y$ and $xy$ are coprime.

So if $x+y|dxy$, then $x+y|d$, so $d\geq x+y$

-1

I think you can explain everithing using this observation

if a prime $p$ divides $a+b$ then $p$ divides $a$ or $b$ (WLOG $p|a$). At this point you $p|(a+b)$ and $p|a$ but then $p|b$ wich means $p|gcd(a,b)$.

That said if you put $d=gcd(a,b)$ and $a=da'$, $b=db'$ you have $a+b|d^2a'b'$ but from the observation $a+b|d^2$ and from this the thesis follow.

Luca
  • 154